Ch. 3. Pulsed and Water ‐ Cooled Magnets T. J. Dolan Magnetic field calculations Coil forces Coil forces RLC circuit equations Distribution of J and B Energy storage E t Switching and transmission Magnetic flux compression Component reliability Power and cooling requirements Coil design considerations Coil design considerations Dolan, IPR 2010 1
Background Chinese Discovered magnetism and invented compass. Mapped the world using compass and astronomical navigation. Zheng He brought knowledge to Europe in 1434 Oersted deflection of compass by current in wire Ampere interaction of current carrying wires Ampere interaction of current-carrying wires Faraday magnetic induction Maxwell equations of electromagnetism Dolan, IPR 2010 2
Required magnetic field equ ed ag e c e d T = 2x10 8 K, n = 2x10 20 m -3 , = 0.1 B = 5.9 T Field at coil is larger: B coil = B o (R o /R coil ) o ( coil ) coil o R coil R o = 6 m, R coil = 2.5 m B B coil il = 14 T 14 T R R o Dolan, IPR 2010 3
Advantages of Water-Cooled Magnets Advantages of Water Cooled Magnets Dolan, IPR 2010 4
Dolan, IPR 2010 5
Dolan, IPR 2010 6
Dolan, IPR 2010 7
(eh and fg are equal and opposite.) Dolan, IPR 2010 8
Dolan, IPR 2010 9
Toroidal Magnetic Field Ripple Toroidal Magnetic Field Ripple Dolan, IPR 2010 10
Law of Biot-Savart Dolan, IPR 2010 11
Field from a Circular Current Ring Field from a Circular Current Ring Dolan, IPR 2010 12
Dolan, IPR 2010 13
k2 K(k) E(k) k2 K(k) E(k) k2 K(k) E(k) k2 K(k) E(k) Dolan, IPR 2010 14
Example – Field of Circular Coil Circular coil, a = 0.5 m, I = 100 kA. Find B( r = 0.4 m, z = 0.6 m ) k 2 = 0.683707 K(k) = 2.05215 E(k) = 1.25125 Br = 0.0025 T Bz = 0.027 T Dolan, IPR 2010 15
Coil Forces F = d F = J x B dV F = d F = J x B dV F = d F = I d ℓ x B for thin wires wires B 1 (at I 2 ) = o I 1 /2 r dF/dL = I 2 B 1 = o I 1 I 2 /2 r = I I /2 r dF/dL = I B Dolan, IPR 2010 16
Force between Circular Loops F z = 2 a I 2 B r1 F 2 a I B Dolan, IPR 2010 17
Force between Circular Loops Example: Two coaxial circular coils with a = 1 m, separated by z = 1 m. , p y Find F. k 2 = 0 8 k = 0.8 K(k) = 2.257, E(k) = 1.178, B r = 0.0697 T F = 4.38x10 5 N Dolan, IPR 2010 18
Tensile Stress in Long Solenoid Coil Example Case: Example Case: r 1 = 1 m, ∆ r = 0.2 m B = 10 T. = 212 MPa Yield stress of copper = 280 MPa Dolan, IPR 2010 19
Force on Torsatron Coils Force on Torsatron Coils Dolan, IPR 2010 20
Force Reduced Torsatron Coils Optimum Pitch angle ~ 42 o g R/a c ~ 7 Dolan, IPR 2010 21
TF Coil Design Considerations g TF coil forces tend to: i increase coil radius a c il di decrease major radius R c bend coils (due to interaction with vertical field) Consider stress concentrations fatigue fatigue creep thermal stress TF coils shaped like “D” have lower stress than circular coils TF coils shaped like “D” have lower stress than circular coils. Dolan, IPR 2010 22
Reduction of Field Errors Coil winding accuracy Coil alignment Coil supports to minimize motion Series connection to equalize currents Series connection to equalize currents Stray B fields from current leads Stray B fields from ferrous objects Dolan, IPR 2010 23
Components p Energy storage Energy storage Switches Transmission lines Coils Diagnostics and controls Diagnostics and controls Dolan, IPR 2010 24
RLC Circuit Equations q R = total resistance L L = total inductance t t l i d t L(d 2 q/dt 2 ) +R(dq/dt)+q/C =0 L(d 2 q/dt 2 ) +R(dq/dt)+q/C =0 q(0) = CV o (dq/dt) o = 0 q(t) = CV o e -at [cos t + (a/ )sin t] I (t) = (V o / L) e -at sin t a=R/2L = [(1/LC) – a 2 ] = [(1/LC) a=R/2L a 2 ] Dolan, IPR 2010 25
Current vs Time Current vs. Time “Undercritically Undercritically damped circuit” Dolan, IPR 2010 26
“Crowbar” Switch S 2 Close S1 at t=0 Close S2 at t=t max Dolan, IPR 2010 27
Resistance of Wire, Rod, Plate, Tube Resistance of Wire, Rod, Plate, Tube ℓ R = dx / S d R / S 0 Example: Copper tube r 1 = 0.02 m, r 2 = 0.025 m, ℓ = 3 m = 2x10 -6 Ohm-m 2x10 Ohm m S = (r 2 2 -r 1 2 ) = 0.000707 m 2 R = ℓ / S = 0.00849 Ohm Dolan, IPR 2010 28
Inductance of N-turn Solenoid Inductance of N turn Solenoid Length ℓ , = r 2 /r 1 Radii r 1 and r 2 1 2 Example: N=20, ℓ = 1 m r = 5 m ℓ = 1 m, r 1 = .5 m, r 2 = .8 m = 1.6, = 2, L/N 2 r 1 = 1.2x10 -6 L = 2.4x10 -4 Henry = ℓ /r 1 Dolan, IPR 2010 29
Parallel Plate Transmission Line Parallel Plate Transmission Line L= o S ℓ K sh / h Dolan, IPR 2010 30
Graph of K h vs (s/h) Graph of K sh vs. (s/h) If /h If s/h << 1, 1 then K sh = 1 Dolan, IPR 2010 31
Coaxial Cable or Tubes Coaxial Cable or Tubes L = ℓ ln(b/a)/2 L = o ℓ ln(b/a)/2 Dolan, IPR 2010 32
Distribution of J and B Distribution of J and B J/ t = J/ t J/ 2 J/ B/ t = 2 B/ Assume 2 B ≈ B/ 2 B/ t ≈ B Then B ≈ B/ 2 ≈ Skin depth = (2/ ) 1/2 th Ski d (2/ ) 1/2 In copper at 1 MHz, = 0.07 mm pp Dolan, IPR 2010 33
Structural Support of Coil Structural Support of Coil Dolan, IPR 2010 34
Distribution of J and B in coil Distribution of J and B in coil Actual Approximate Dolan, IPR 2010 35
Axial Distribution of B in Solenoids Single-turn Uniform J Dolan, IPR 2010 36
Coil Melting and Yielding Yielding at B ~ B y [(r 2 -r 1 )/2r 1 ] 1/2 B (Cu SS) = 25 T B y (Cu, SS) = 25 T B y (Ta) = 32 T B (Ta) = 32 T Melting at B ~ B / 1/2 Melting at B ~ B e / 1/2 ~ 3 ~ 3 B e (Cu, SS) = 90 T B e (Ta) = 137 T Fatigue failures after many shots Sudden B > 70 T Coil explodes Dolan, IPR 2010 37
Energy Storage Systems Dolan, IPR 2010 38
Scyllac Capacitor Scyllac Capacitor 60 kV, 1.85 F Dolan, IPR 2010 39
Energy Storage System Costs Energy Storage System Costs Fusion experiments Power grids S l Solar power Wind power flywheel ~1980 values Dolan, IPR 2010 40
Inductive Energy Storage Opening switch S1 forces current to flow through plasma confinement coil. S 1 : difficult to prevent arcing Transfer efficiency = L s L/(L s +L) 2 25% Dolan, IPR 2010 41
50 MJ Homopolar Generator E r = v xB z U of Texas U. of Texas Dolan, IPR 2010 42
Flywheel Energy Storage Can store about 500 MJ/m 3 Can store about 500 MJ/m Princeton Plasma Physics Laboratory Princeton Plasma Physics Laboratory motor-generator 200 MW, 3 s 200 MW 3 More expensive than homopolar system Dolan, IPR 2010 43
Spark Gap Switch p p Dolan, IPR 2010 44
Los Alamos Dual Spark Gap Switch Low “jitter”: 3240 switches fired within 10 ns. Dolan, IPR 2010 45
Exploding Foil Switch Dolan, IPR 2010 46
Marx Bank +400 kV 100 kV 100 kV 100 kV 100 kV Capacitors are charged in parallel Capacitors are charged in parallel Then discharged in series to give high voltage Dolan, IPR 2010 47
High Voltage Coaxial Cable Scyllac experiment had 250 km of these cables Scyllac experiment had 250 km of these cables. 10 5 shot reliability. Dolan, IPR 2010 48
Magnetic Flux Compression “Bellows” type Dolan, IPR 2010 49
Imploding Metallic Liner Imploding Metallic Liner explosive liner Debris Debris low Liner flux Compressed flux Dolan, IPR 2010 50
Failure Rates f j dt = failure probability of item j during dt t Failure probability between 0 and t: p = dt f(t) Failure probability between 0 and t: p f = dt f j (t) 0 Probability of not failing before time t = 1 p Probability of not failing before time t = 1-p f “Failure Rate” at time t: r j (t) = p f /(1-p f ) Total failure rate r(t) = r j (t) (failures per second) j Estimated time between failures ETBF = 1/r(t) Dolan, IPR 2010 51
Component Reliability Capacitors, cables, automobiles automobiles, … Dolan, IPR 2010 52
Estimated Time to Next Failure Example: 100 capacitors with j = 10 -4 per shot and 600 cables with j = 2x10 -4 per shot Find ETNF Find ETNF ETNF = 1 / [100x10 -4 + 600x2x10 -4 ] = 7.7 shots Similar analysis for laser systems, automobiles, etc. Dolan, IPR 2010 53
Coil Power Requirements = (copper volume)/(coil volume) dP = J c 2 dV Circular coils: P = J c 2 dz 2 rdr over coil volume Field at center of solenoid with length L, radii r 1 and r 2 : f B z = 3/2 o g( )(P/ r 1 ) 1/2 o g( )( 1 ) z where = r 2 /r 1 = L/2r 1 Dolan, IPR 2010 54
Recommend
More recommend
